Question

Recall that a function f is even if f(−x)=f(x) and odd if f(−x)=−f(x), for all x in the domain of f. Assuming that f is differentiable, prove: (a) f′ is odd if f is even (b) f′ is even if f is odd.

Answer 1

(a) To prove that if f is even, then f' is odd, we need to show that f'(-x) = -f'(x) for all x in the domain of f.

Assuming f is even, we know that f(-x) = f(x) for all x in the domain of f.

Now, we can calculate the derivative of f:

f'(-x) = lim(h->0) [ f(-x - h) - f(-x) ] / h
= lim(h->0) [ f(x + h) - f(x) ] / h (using f(-x) = f(x))
= f'(x)

Similarly,

-f'(x) = - lim(h->0) [ f(x + h) - f(x) ] / h
= lim(h->0) [ -(f(x + h) - f(x)) ] / h
= lim(h->0) [ f(x) - f(x + h) ] / h
= lim(h->0) [ f(-x - h) - f(-x) ] / h (using f(-x) = f(x))
= f'(-x)

Therefore, we have shown that f'(-x) = -f'(x), which means f' is odd if f is even.

(b) To prove that if f is odd, then f' is even, we need to show that f'(-x) = f'(x) for all x in the domain of f.

Assuming f is odd, we know that f(-x) = -f(x) for all x in the domain of f.

Now, let's calculate the derivative of f:

f'(-x) = lim(h->0) [ f(-x - h) - f(-x) ] / h
= lim(h->0) [ -f(x + h) - (-f(x)) ] / h (using f(-x) = -f(x))
= lim(h->0) [ -f(x + h) + f(x) ] / h
= - lim(h->0) [ f(x + h) - f(x) ] / h
= - f'(x)

Therefore, we have shown that f'(-x) = f'(x), which means f' is even if f is odd.

By proving both (a) and (b), we have established that if f is even, then f' is odd, and if f is odd, then f' is even.

If you have any more questions, feel free to ask!